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Binomial Coefficients

If the coeffi...

Question

If the coefficient of $x^{r - 1}$ , $x^{r}$ , $x^{r + 1}$ in the expansion of ${1 + x}^{n}$ are in AP.then $n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0$

True

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False

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Solution

The correct option is A True

$Given, x^{r - 1}, x^{r}, x^{r + 1} in the expansion$
$of (1 + x)^{n} are in A P$
$T_{r + 1} =^{n} C_{r} x^{r}$
$T_{r} =^{n} C_{r - 1} x^{r - 1}$
$T_{r + 2} =^{n} C_{r + 1} x^{r + 1}$

Given, $^{n} C_{r - 1},^{n} C_{r},^{n} C_{r + 1}$ are in AP

$\Rightarrow^{n} C_{r - 1} +^{n} C_{r + 1} = 2^{n} C_{r}$

$\Rightarrow \frac{^{n} c_{r - 1}}{^{n} c_{r}} + \frac{^{n} c_{r + 1}}{^{n} c_{r}} = 2$

${\begin{matrix} n_{C_{r}} = \frac{n!}{r! (n - r)!}} \end{matrix}$

$\Rightarrow \frac{r}{n - r + 1} + \frac{n - r}{r + 1} = 2$

$\Rightarrow r^{2} + r + (n - r)^{2} + n - r = 2 (n - r + 1) (r + 1)$

$\Rightarrow r^{2} + n^{2} + r^{2} - 2 n r + n = 2 n r + 2 n - 2 r^{2} - 2 r + 2 r + 2$

$\Rightarrow 4 r^{2} - 4 n r + n^{2} = n + 2$

$\Rightarrow (2 r - n)^{2} = n + 2$ so $4 r^{2} - (4 r + 1) n + n^{2} - 2 = 0$

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Similar questions

If the coefficients of $x^{r - 1}, x^{r}$ and $x^{r + 1}$ in the binomial expansion of $(1 + x)^{n}$ are in AP, prove that
$n^{2} - n (4 r + 1) + 4 r^{2} - 2 = 0$

Q. If

x^{r}

occurs in the expansion of

{(x + \frac{1}{x})}^{n}

, find its coefficient.

Q. If the coefficient of

r^{t h}, (r + 1)^{t h}

and

(r + 2)^{t h}

terms in the expansion of

(1 + x)^{n}

are in A.P., then show that

n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0

Q. Consider the expansion of

(1 + x)^{2 n + 1}

If the coefficients of

x^{r}

and

x^{r + 1}

are equal in the expansion, then

r

is equal to

Q. If

C_{r}

is coefficient of

x^{r}

in expansion of

{(1 + x)}^{n}, n ϵ N, n \geq 4

then

n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}}

is divisible (remainder 0) by
(1+x)

^{n}

, n

ϵ N

, n

\geq 4

c_{r}, x^{r}

n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}} = 0

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If the coefficient of xr−1, xr,xr+1 in the expansion of 1+xn are in AP.then n2−(4r+1)n+4r2−2=0

If the coefficient of $x^{r - 1}$ , $x^{r}$ , $x^{r + 1}$ in the expansion of ${1 + x}^{n}$ are in AP.then $n^{2} - (4 r + 1) n + 4 r^{2} - 2 = 0$